a)

`1/1-1/2`

`=2/2-1/2`

`=1/2`

b)

`1/(1*2)+1/(2*3)`

`=1/1-1/2+1/2-1/3`

`=1/1-1/3`

`=3/3-1/3`

`=2/3`

c)

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{1}-\dfrac{1}{100}\\ =\dfrac{99}{100}\)

d) 

\(\dfrac{3}{1\cdot2}+\dfrac{3}{2\cdot3}+...+\dfrac{3}{99\cdot100}\) đề phải như thế này chứ nhỉ?

\(=\dfrac{1\cdot3}{1\cdot2}+\dfrac{1\cdot3}{2\cdot3}+...+\dfrac{1\cdot3}{99\cdot100}\\ =3\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\\ =3\cdot\dfrac{99}{100}\\ =\dfrac{297}{100}\)

b: Tổng của N là:

\(\dfrac{49\cdot48}{2}=49\cdot24=1176\)

chào nick thứ 2 đây

a) \(3M=1.2.3+2.3.3+...+48.49.3=1.2.3+2.3.\left(4-1\right)+...+48.49.\left(50-47\right)=1.2.3+2.3.4-1.2.3+...+48.49.50-47.48.49=48.49.50\Rightarrow M=\dfrac{48.49.50}{3}\Rightarrow M=39200\)

b) Tương tự câu a

Program i: integer;

s: longint;

Begin

s:=0;

for i:=1 to 99 do s:=s + i*(i+1);

write('S=',s);

readln

end.

uses crt;

var i,s:longint;

begin

clrscr;

s:=0;

for i:=1 to 99 do 

  s:=s+i*(i+1);

writeln('S=',s);

readln;

end.

Tk:
Đặt P = 1.2+2.3+3.4+...+99.100

3P = 1.2.3+2.3.3+3.4.3+...+99.100+3

3P = 1.2 (3-0) +2.3(4-1)+3.4(5-2) +...+ 99.100( 101-98)

3P = ( 1.2.3 + 2.3.4 + 3.4.5 + 99.100.101 ) -( 0.1.2 + 1.2.3 + 2.3.4 + ....+ 98.99.100)

3P = 99.100.101 - 0.1.2

3P = 999900 - 0

3P = 999900

P = 999900 : 3

P = 333300

\(A=1.2+2.3+3.4+...+99.100\)

\(\Rightarrow3A=1.2.3+2.3.3+...+99.100.3\)

\(=1.2.3+2.3.\left(4-1\right)+3.4\left(5-2\right)+...+99.100\left(101-98\right)\)

\(=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-....-98.99.100+99.100.101\)

\(=99.100.101\)

\(\Rightarrow A=\dfrac{99.100.101}{3}=333300\)

=1x2+2x3+3x4+...+99x100

kho qua khong bit

Gọi A là biểu thức ta có: 
A = 1.2+2.3+3.4+......+99.100 
Gấp A lên 3 lần ta có: 
A . 3 = 1.2.3 + 2.3.3 + 3.4.3 + … + 99.100.3 
A . 3 = 1.2.3 + 2.3.(4 - 1) + 3.4.( 5 - 2) + … + 99.100. (101 - 98) 
A . 3 = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + … + 99.100.101 - 98.99.100 
A . 3 = 99.100.101 
A = 99.100.101 : 3 
A = 33.100.101 
A = 333 300

Đặt biểu thức là A ta có:

A=1.2+2.3+3.4+...+99.100 (1)

Nhân 2 vế của đẳng thức (1) với 3 ta được:

3A=3.(1.2+2.3+3.4+4.5+...+99.100)

3A=1.2.3+2.3.3+3.3.4+4.5.3+...+99.100.3

3A=1.2.3+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+...+99.100.(101-98)

3A=1.2.3+2.3.4-2.3.1+3.4.5-2.3.4+4.5.6-4.5.3+...+99.100.101-99.100.98

3A=99.100.101

3A=999900

A=999900:3=333300

1.2+2.3+3.4+...+99.100

=(99.100.101-0.1.2):3=333300

A = 1.2+2.3+3.4+......+99.100 
Gấp A lên 3 lần ta có: 
A . 3 = 1.2.3 + 2.3.3 + 3.4.3 + … + 99.100.3 
A . 3 = 1.2.3 + 2.3.(4 - 1) + 3.4.( 5 - 2) + … + 99.100. (101 - 98) 
A . 3 = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + … + 99.100.101 - 98.99.100 
A . 3 = 99.100.101 
A = 99.100.101 : 3 
A = 33.100.101 
A = 333 300

Đặt A = 1.2 + 2.3 + 3.4 + ...+99.100

=> 3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 99.100.3

=> 3A = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) + ... + 99.100.(101-98)

=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + 4.5.6 - 3.4.5 + ... + 99.100.101-98.99.100

=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + 3.4.5 - 3.4.5 + ... + 99.100.101

=> 3A = 99.100.101

=> 3A = 999900

=> A = 999900 : 3

=> A = 333300

 Vậy A = 333300